Navier-Stokes Formulation












2












$begingroup$


In some artictle I've found Navier-Stokes Momentum Equation written in a following form
$$frac{partial u}{partial t}+text{div}(uotimes u)+nabla p - nuDelta u =f$$
Since usually it appears with $ucdot nabla u $ term instead of $text{div}(uotimes u)$. It all came up from using following form of the Newton Law $$underbrace{frac{d}{dt}int_B(rho u)(t,cdot)dx}_{text{change of the linear momentum}}=underbrace{-int_{partial B} (rho uotimes u )(t,cdot)n(cdot)dS}_{text{flux of the momentum through boundary}}+underbrace{F_B}_{text{applied force}}\
F_B=underbrace{int_B(rho f)(t,cdot)dx}_{text{volume forces}}+underbrace{int_{partial B}text{T}(t,cdot)n(cdot)dS}_{text{surface forces (the tension)}}$$

Then we end up with integral form
$$int_BBig(frac{partial}{partial t}(rho u)(t,cdot)+text{div}(rho uotimes u)(t,cdot)-(rho f)(t,cdot)-text{divT}(t,cdot)Big) dx=0.$$
Defining stress tensor $T$ and assuming constant density we end up with the first equation, but since I'm not really fluent with tensors I could use some help understanding the difference between this formulation and the original one, and how to make calculations to get one form another.



Thank you!










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    In some artictle I've found Navier-Stokes Momentum Equation written in a following form
    $$frac{partial u}{partial t}+text{div}(uotimes u)+nabla p - nuDelta u =f$$
    Since usually it appears with $ucdot nabla u $ term instead of $text{div}(uotimes u)$. It all came up from using following form of the Newton Law $$underbrace{frac{d}{dt}int_B(rho u)(t,cdot)dx}_{text{change of the linear momentum}}=underbrace{-int_{partial B} (rho uotimes u )(t,cdot)n(cdot)dS}_{text{flux of the momentum through boundary}}+underbrace{F_B}_{text{applied force}}\
    F_B=underbrace{int_B(rho f)(t,cdot)dx}_{text{volume forces}}+underbrace{int_{partial B}text{T}(t,cdot)n(cdot)dS}_{text{surface forces (the tension)}}$$

    Then we end up with integral form
    $$int_BBig(frac{partial}{partial t}(rho u)(t,cdot)+text{div}(rho uotimes u)(t,cdot)-(rho f)(t,cdot)-text{divT}(t,cdot)Big) dx=0.$$
    Defining stress tensor $T$ and assuming constant density we end up with the first equation, but since I'm not really fluent with tensors I could use some help understanding the difference between this formulation and the original one, and how to make calculations to get one form another.



    Thank you!










    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      In some artictle I've found Navier-Stokes Momentum Equation written in a following form
      $$frac{partial u}{partial t}+text{div}(uotimes u)+nabla p - nuDelta u =f$$
      Since usually it appears with $ucdot nabla u $ term instead of $text{div}(uotimes u)$. It all came up from using following form of the Newton Law $$underbrace{frac{d}{dt}int_B(rho u)(t,cdot)dx}_{text{change of the linear momentum}}=underbrace{-int_{partial B} (rho uotimes u )(t,cdot)n(cdot)dS}_{text{flux of the momentum through boundary}}+underbrace{F_B}_{text{applied force}}\
      F_B=underbrace{int_B(rho f)(t,cdot)dx}_{text{volume forces}}+underbrace{int_{partial B}text{T}(t,cdot)n(cdot)dS}_{text{surface forces (the tension)}}$$

      Then we end up with integral form
      $$int_BBig(frac{partial}{partial t}(rho u)(t,cdot)+text{div}(rho uotimes u)(t,cdot)-(rho f)(t,cdot)-text{divT}(t,cdot)Big) dx=0.$$
      Defining stress tensor $T$ and assuming constant density we end up with the first equation, but since I'm not really fluent with tensors I could use some help understanding the difference between this formulation and the original one, and how to make calculations to get one form another.



      Thank you!










      share|cite|improve this question









      $endgroup$




      In some artictle I've found Navier-Stokes Momentum Equation written in a following form
      $$frac{partial u}{partial t}+text{div}(uotimes u)+nabla p - nuDelta u =f$$
      Since usually it appears with $ucdot nabla u $ term instead of $text{div}(uotimes u)$. It all came up from using following form of the Newton Law $$underbrace{frac{d}{dt}int_B(rho u)(t,cdot)dx}_{text{change of the linear momentum}}=underbrace{-int_{partial B} (rho uotimes u )(t,cdot)n(cdot)dS}_{text{flux of the momentum through boundary}}+underbrace{F_B}_{text{applied force}}\
      F_B=underbrace{int_B(rho f)(t,cdot)dx}_{text{volume forces}}+underbrace{int_{partial B}text{T}(t,cdot)n(cdot)dS}_{text{surface forces (the tension)}}$$

      Then we end up with integral form
      $$int_BBig(frac{partial}{partial t}(rho u)(t,cdot)+text{div}(rho uotimes u)(t,cdot)-(rho f)(t,cdot)-text{divT}(t,cdot)Big) dx=0.$$
      Defining stress tensor $T$ and assuming constant density we end up with the first equation, but since I'm not really fluent with tensors I could use some help understanding the difference between this formulation and the original one, and how to make calculations to get one form another.



      Thank you!







      tensors fluid-dynamics






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      share|cite|improve this question











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      asked Dec 18 '18 at 14:51









      user396656user396656

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      232






















          1 Answer
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          $begingroup$

          The most generic form of the Navier-Stokes equation is
          $$
          frac{partial (rho mathbf{u})}{partial t} + nabla cdot(rho mathbf{u} otimes mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{1}
          $$

          in which $mathbf{S}$ is the shear stress tensor ($mathbf{S}=munabla mathbf{u}$ in your case). The continuity equation is
          $$
          frac{partial rho}{partial t}+nabla cdot (rho mathbf{u})=0.
          tag{2}
          $$

          Using the product rule for derivatives in equation $(1)$ leads to
          $$
          rho frac{partial mathbf{u}}{partial t} + mathbf{u}frac{partial rho}{partial t} + rho mathbf{u} cdot nabla mathbf{u} + mathbf{u} nabla cdot(rho mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{3}
          $$

          and using the continuity equation we see that the second and fourth terms in LHS cancel each other, leading to
          $$
          rholeft( frac{partial mathbf{u}}{partial t} + mathbf{u} cdot nabla mathbf{u} right) = nabla p - mathbf{f} + nabla cdot mathbf{S}.
          tag{4}
          $$

          Since it's usually assumed that the continuity equation holds, equation $(4)$ is completly equivalent to equation $(1)$. In practice, however, there are some "differences":




          1. Equation $(1)$ is called the conservative form of Navier-Stokes equation, while equation $(4)$ is called the non-conservative form. These names are a bit misleading: both equations are the momentum conservation equations. However, when solving numerically the governing equations of fluid dynamics, it's sometimes more useful to use equation $(1)$. It's basically due to the fact that accross a shock wave the velocity $mathbf{u}$ is discontinuous (and, therefore, equation $(4)$ has the gradient of a discontinuous function), while $rho mathbf{u} otimes mathbf{u}$ is continuous even accross the shock. See that equation $(1)$ is called conservative form because it has derivatives of a conserved quantity (the momentum, i.e., $rho mathbf{u}$) while equation $(4)$ has derivatives of a non-conserved quantity (the velocity).


          2. Equation $(4)$ explicitly shows the transport of momentum in the term $mathbf{u} cdot nabla mathbf{u}$. Notice that the conservation equation of the property $phi$ (which can be enthalpy, vorticity, chemical species, etc.) will have a term $mathbf{u} cdot nabla phi$. Therefore, equation $(4)$ "looks like" every other conservation equation. It's usually defined the material derivative of property $phi$ as
            $$
            frac{D phi}{Dt} = frac{partialphi}{partial t} + mathbf{u} cdot nablaphi,
            $$

            which can be interpreted as the rate of change of the property $phi$ along time in a particle of fluid while this particle is transported by the flow. Then, the conservation equation of any property $phi$ can be written generically as
            $$
            frac{D phi}{Dt} = text{source terms},
            $$

            in which the source terms for the case $phi=mathbf{u}$ (i.e., the Navier-Stokes equation) are $(nabla p - mathbf{f} + nabla cdot mathbf{S})/rho$.



          Summarizing: both equations are the same. When you need to solve them numerically, equation $(1)$ can be more suitable. If you want to interpret the physical meaning of the terms of Navier-Stokes equation, equation $(4)$ is more suitable.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, really well explained!
            $endgroup$
            – user396656
            Dec 18 '18 at 16:06











          Your Answer





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          $begingroup$

          The most generic form of the Navier-Stokes equation is
          $$
          frac{partial (rho mathbf{u})}{partial t} + nabla cdot(rho mathbf{u} otimes mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{1}
          $$

          in which $mathbf{S}$ is the shear stress tensor ($mathbf{S}=munabla mathbf{u}$ in your case). The continuity equation is
          $$
          frac{partial rho}{partial t}+nabla cdot (rho mathbf{u})=0.
          tag{2}
          $$

          Using the product rule for derivatives in equation $(1)$ leads to
          $$
          rho frac{partial mathbf{u}}{partial t} + mathbf{u}frac{partial rho}{partial t} + rho mathbf{u} cdot nabla mathbf{u} + mathbf{u} nabla cdot(rho mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{3}
          $$

          and using the continuity equation we see that the second and fourth terms in LHS cancel each other, leading to
          $$
          rholeft( frac{partial mathbf{u}}{partial t} + mathbf{u} cdot nabla mathbf{u} right) = nabla p - mathbf{f} + nabla cdot mathbf{S}.
          tag{4}
          $$

          Since it's usually assumed that the continuity equation holds, equation $(4)$ is completly equivalent to equation $(1)$. In practice, however, there are some "differences":




          1. Equation $(1)$ is called the conservative form of Navier-Stokes equation, while equation $(4)$ is called the non-conservative form. These names are a bit misleading: both equations are the momentum conservation equations. However, when solving numerically the governing equations of fluid dynamics, it's sometimes more useful to use equation $(1)$. It's basically due to the fact that accross a shock wave the velocity $mathbf{u}$ is discontinuous (and, therefore, equation $(4)$ has the gradient of a discontinuous function), while $rho mathbf{u} otimes mathbf{u}$ is continuous even accross the shock. See that equation $(1)$ is called conservative form because it has derivatives of a conserved quantity (the momentum, i.e., $rho mathbf{u}$) while equation $(4)$ has derivatives of a non-conserved quantity (the velocity).


          2. Equation $(4)$ explicitly shows the transport of momentum in the term $mathbf{u} cdot nabla mathbf{u}$. Notice that the conservation equation of the property $phi$ (which can be enthalpy, vorticity, chemical species, etc.) will have a term $mathbf{u} cdot nabla phi$. Therefore, equation $(4)$ "looks like" every other conservation equation. It's usually defined the material derivative of property $phi$ as
            $$
            frac{D phi}{Dt} = frac{partialphi}{partial t} + mathbf{u} cdot nablaphi,
            $$

            which can be interpreted as the rate of change of the property $phi$ along time in a particle of fluid while this particle is transported by the flow. Then, the conservation equation of any property $phi$ can be written generically as
            $$
            frac{D phi}{Dt} = text{source terms},
            $$

            in which the source terms for the case $phi=mathbf{u}$ (i.e., the Navier-Stokes equation) are $(nabla p - mathbf{f} + nabla cdot mathbf{S})/rho$.



          Summarizing: both equations are the same. When you need to solve them numerically, equation $(1)$ can be more suitable. If you want to interpret the physical meaning of the terms of Navier-Stokes equation, equation $(4)$ is more suitable.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, really well explained!
            $endgroup$
            – user396656
            Dec 18 '18 at 16:06
















          0












          $begingroup$

          The most generic form of the Navier-Stokes equation is
          $$
          frac{partial (rho mathbf{u})}{partial t} + nabla cdot(rho mathbf{u} otimes mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{1}
          $$

          in which $mathbf{S}$ is the shear stress tensor ($mathbf{S}=munabla mathbf{u}$ in your case). The continuity equation is
          $$
          frac{partial rho}{partial t}+nabla cdot (rho mathbf{u})=0.
          tag{2}
          $$

          Using the product rule for derivatives in equation $(1)$ leads to
          $$
          rho frac{partial mathbf{u}}{partial t} + mathbf{u}frac{partial rho}{partial t} + rho mathbf{u} cdot nabla mathbf{u} + mathbf{u} nabla cdot(rho mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{3}
          $$

          and using the continuity equation we see that the second and fourth terms in LHS cancel each other, leading to
          $$
          rholeft( frac{partial mathbf{u}}{partial t} + mathbf{u} cdot nabla mathbf{u} right) = nabla p - mathbf{f} + nabla cdot mathbf{S}.
          tag{4}
          $$

          Since it's usually assumed that the continuity equation holds, equation $(4)$ is completly equivalent to equation $(1)$. In practice, however, there are some "differences":




          1. Equation $(1)$ is called the conservative form of Navier-Stokes equation, while equation $(4)$ is called the non-conservative form. These names are a bit misleading: both equations are the momentum conservation equations. However, when solving numerically the governing equations of fluid dynamics, it's sometimes more useful to use equation $(1)$. It's basically due to the fact that accross a shock wave the velocity $mathbf{u}$ is discontinuous (and, therefore, equation $(4)$ has the gradient of a discontinuous function), while $rho mathbf{u} otimes mathbf{u}$ is continuous even accross the shock. See that equation $(1)$ is called conservative form because it has derivatives of a conserved quantity (the momentum, i.e., $rho mathbf{u}$) while equation $(4)$ has derivatives of a non-conserved quantity (the velocity).


          2. Equation $(4)$ explicitly shows the transport of momentum in the term $mathbf{u} cdot nabla mathbf{u}$. Notice that the conservation equation of the property $phi$ (which can be enthalpy, vorticity, chemical species, etc.) will have a term $mathbf{u} cdot nabla phi$. Therefore, equation $(4)$ "looks like" every other conservation equation. It's usually defined the material derivative of property $phi$ as
            $$
            frac{D phi}{Dt} = frac{partialphi}{partial t} + mathbf{u} cdot nablaphi,
            $$

            which can be interpreted as the rate of change of the property $phi$ along time in a particle of fluid while this particle is transported by the flow. Then, the conservation equation of any property $phi$ can be written generically as
            $$
            frac{D phi}{Dt} = text{source terms},
            $$

            in which the source terms for the case $phi=mathbf{u}$ (i.e., the Navier-Stokes equation) are $(nabla p - mathbf{f} + nabla cdot mathbf{S})/rho$.



          Summarizing: both equations are the same. When you need to solve them numerically, equation $(1)$ can be more suitable. If you want to interpret the physical meaning of the terms of Navier-Stokes equation, equation $(4)$ is more suitable.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Thank you very much, really well explained!
            $endgroup$
            – user396656
            Dec 18 '18 at 16:06














          0












          0








          0





          $begingroup$

          The most generic form of the Navier-Stokes equation is
          $$
          frac{partial (rho mathbf{u})}{partial t} + nabla cdot(rho mathbf{u} otimes mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{1}
          $$

          in which $mathbf{S}$ is the shear stress tensor ($mathbf{S}=munabla mathbf{u}$ in your case). The continuity equation is
          $$
          frac{partial rho}{partial t}+nabla cdot (rho mathbf{u})=0.
          tag{2}
          $$

          Using the product rule for derivatives in equation $(1)$ leads to
          $$
          rho frac{partial mathbf{u}}{partial t} + mathbf{u}frac{partial rho}{partial t} + rho mathbf{u} cdot nabla mathbf{u} + mathbf{u} nabla cdot(rho mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{3}
          $$

          and using the continuity equation we see that the second and fourth terms in LHS cancel each other, leading to
          $$
          rholeft( frac{partial mathbf{u}}{partial t} + mathbf{u} cdot nabla mathbf{u} right) = nabla p - mathbf{f} + nabla cdot mathbf{S}.
          tag{4}
          $$

          Since it's usually assumed that the continuity equation holds, equation $(4)$ is completly equivalent to equation $(1)$. In practice, however, there are some "differences":




          1. Equation $(1)$ is called the conservative form of Navier-Stokes equation, while equation $(4)$ is called the non-conservative form. These names are a bit misleading: both equations are the momentum conservation equations. However, when solving numerically the governing equations of fluid dynamics, it's sometimes more useful to use equation $(1)$. It's basically due to the fact that accross a shock wave the velocity $mathbf{u}$ is discontinuous (and, therefore, equation $(4)$ has the gradient of a discontinuous function), while $rho mathbf{u} otimes mathbf{u}$ is continuous even accross the shock. See that equation $(1)$ is called conservative form because it has derivatives of a conserved quantity (the momentum, i.e., $rho mathbf{u}$) while equation $(4)$ has derivatives of a non-conserved quantity (the velocity).


          2. Equation $(4)$ explicitly shows the transport of momentum in the term $mathbf{u} cdot nabla mathbf{u}$. Notice that the conservation equation of the property $phi$ (which can be enthalpy, vorticity, chemical species, etc.) will have a term $mathbf{u} cdot nabla phi$. Therefore, equation $(4)$ "looks like" every other conservation equation. It's usually defined the material derivative of property $phi$ as
            $$
            frac{D phi}{Dt} = frac{partialphi}{partial t} + mathbf{u} cdot nablaphi,
            $$

            which can be interpreted as the rate of change of the property $phi$ along time in a particle of fluid while this particle is transported by the flow. Then, the conservation equation of any property $phi$ can be written generically as
            $$
            frac{D phi}{Dt} = text{source terms},
            $$

            in which the source terms for the case $phi=mathbf{u}$ (i.e., the Navier-Stokes equation) are $(nabla p - mathbf{f} + nabla cdot mathbf{S})/rho$.



          Summarizing: both equations are the same. When you need to solve them numerically, equation $(1)$ can be more suitable. If you want to interpret the physical meaning of the terms of Navier-Stokes equation, equation $(4)$ is more suitable.






          share|cite|improve this answer









          $endgroup$



          The most generic form of the Navier-Stokes equation is
          $$
          frac{partial (rho mathbf{u})}{partial t} + nabla cdot(rho mathbf{u} otimes mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{1}
          $$

          in which $mathbf{S}$ is the shear stress tensor ($mathbf{S}=munabla mathbf{u}$ in your case). The continuity equation is
          $$
          frac{partial rho}{partial t}+nabla cdot (rho mathbf{u})=0.
          tag{2}
          $$

          Using the product rule for derivatives in equation $(1)$ leads to
          $$
          rho frac{partial mathbf{u}}{partial t} + mathbf{u}frac{partial rho}{partial t} + rho mathbf{u} cdot nabla mathbf{u} + mathbf{u} nabla cdot(rho mathbf{u}) = nabla p - mathbf{f} + nabla cdot mathbf{S},
          tag{3}
          $$

          and using the continuity equation we see that the second and fourth terms in LHS cancel each other, leading to
          $$
          rholeft( frac{partial mathbf{u}}{partial t} + mathbf{u} cdot nabla mathbf{u} right) = nabla p - mathbf{f} + nabla cdot mathbf{S}.
          tag{4}
          $$

          Since it's usually assumed that the continuity equation holds, equation $(4)$ is completly equivalent to equation $(1)$. In practice, however, there are some "differences":




          1. Equation $(1)$ is called the conservative form of Navier-Stokes equation, while equation $(4)$ is called the non-conservative form. These names are a bit misleading: both equations are the momentum conservation equations. However, when solving numerically the governing equations of fluid dynamics, it's sometimes more useful to use equation $(1)$. It's basically due to the fact that accross a shock wave the velocity $mathbf{u}$ is discontinuous (and, therefore, equation $(4)$ has the gradient of a discontinuous function), while $rho mathbf{u} otimes mathbf{u}$ is continuous even accross the shock. See that equation $(1)$ is called conservative form because it has derivatives of a conserved quantity (the momentum, i.e., $rho mathbf{u}$) while equation $(4)$ has derivatives of a non-conserved quantity (the velocity).


          2. Equation $(4)$ explicitly shows the transport of momentum in the term $mathbf{u} cdot nabla mathbf{u}$. Notice that the conservation equation of the property $phi$ (which can be enthalpy, vorticity, chemical species, etc.) will have a term $mathbf{u} cdot nabla phi$. Therefore, equation $(4)$ "looks like" every other conservation equation. It's usually defined the material derivative of property $phi$ as
            $$
            frac{D phi}{Dt} = frac{partialphi}{partial t} + mathbf{u} cdot nablaphi,
            $$

            which can be interpreted as the rate of change of the property $phi$ along time in a particle of fluid while this particle is transported by the flow. Then, the conservation equation of any property $phi$ can be written generically as
            $$
            frac{D phi}{Dt} = text{source terms},
            $$

            in which the source terms for the case $phi=mathbf{u}$ (i.e., the Navier-Stokes equation) are $(nabla p - mathbf{f} + nabla cdot mathbf{S})/rho$.



          Summarizing: both equations are the same. When you need to solve them numerically, equation $(1)$ can be more suitable. If you want to interpret the physical meaning of the terms of Navier-Stokes equation, equation $(4)$ is more suitable.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 16:00









          rafa11111rafa11111

          1,1242417




          1,1242417








          • 1




            $begingroup$
            Thank you very much, really well explained!
            $endgroup$
            – user396656
            Dec 18 '18 at 16:06














          • 1




            $begingroup$
            Thank you very much, really well explained!
            $endgroup$
            – user396656
            Dec 18 '18 at 16:06








          1




          1




          $begingroup$
          Thank you very much, really well explained!
          $endgroup$
          – user396656
          Dec 18 '18 at 16:06




          $begingroup$
          Thank you very much, really well explained!
          $endgroup$
          – user396656
          Dec 18 '18 at 16:06


















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